Question

graph the circle $x^2 + y^2 + 10x + 6y + 18 = 0$.

Explanation:

Step1: Rewrite the equation in standard form

The general equation of a circle is \(x^{2}+y^{2}+Dx + Ey+F = 0\), and the standard form is \((x - h)^{2}+(y - k)^{2}=r^{2}\), where \((h,k)\) is the center and \(r\) is the radius. We complete the square for \(x\) and \(y\) terms.

For the given equation \(x^{2}+y^{2}+10x + 6y+18 = 0\), group \(x\) terms and \(y\) terms:
\((x^{2}+10x)+(y^{2}+6y)=- 18\)

Complete the square for \(x\): take half of \(10\) (which is \(5\)), square it (\(25\)), so add and subtract \(25\) for \(x\) terms.
Complete the square for \(y\): take half of \(6\) (which is \(3\)), square it (\(9\)), so add and subtract \(9\) for \(y\) terms.

\((x^{2}+10x + 25-25)+(y^{2}+6y + 9-9)=-18\)

\((x + 5)^{2}-25+(y + 3)^{2}-9=-18\)

Step2: Simplify to get standard form

\((x + 5)^{2}+(y + 3)^{2}-34=-18\)

\((x + 5)^{2}+(y + 3)^{2}=34 - 18\)

\((x + 5)^{2}+(y + 3)^{2}=16\)

So the center of the circle is \((-5,-3)\) and the radius \(r = \sqrt{16}=4\).

To graph the circle:

• Plot the center \((-5,-3)\) on the coordinate plane.
• Since the radius is \(4\), move \(4\) units up, down, left, and right from the center to get points on the circle:
• Up: \((-5,-3 + 4)=(-5,1)\)
• Down: \((-5,-3-4)=(-5,-7)\)
• Left: \((-5 - 4,-3)=(-9,-3)\)
• Right: \((-5 + 4,-3)=(-1,-3)\)
• Then draw a circle passing through these points.

Answer:

The circle has center \((-5,-3)\) and radius \(4\), and is graphed by plotting the center and using the radius to find key points on the circle and drawing the circle through them. The standard form of the circle's equation is \((x + 5)^{2}+(y + 3)^{2}=16\).